Formula sheets, the key method steps for every topic and 30 practice problems, with full solutions at the back.
The simply supported beam has a pin at A and a roller at B. Find the vertical support reaction RA.
The simply supported beam has a pin at A and a roller at B. A load P acts at 2.5 m from A, inclined at 60° to the horizontal, and a vertical load Q acts at 4.5 m from A. Find the magnitude of the total reaction force RA at the pin A.
Use the method of joints to find the force FBC in member BC. Give tension as positive and compression as negative.
Use the method of joints to find the force FAC in member AC. Give tension as positive and compression as negative.
Find the bending moment Mx at the section x = 4.5 m from support A. Use sagging moments positive and upward shear on the left face positive.
The marks A to G split the beam into six equal parts. At which mark is the bending moment largest in size, sagging or hogging? Click it in the diagram.
A simply supported beam carries a point load P at a = 2.5 m from support A. Find the deflection δP under the load.
A simply supported timber beam of rectangular section, 100 mm wide, carries a uniformly distributed load w over its whole span. Its midspan deflection may not exceed span/250. Find the smallest depth hreq of the section.
A column of length L made of aluminium is fixed at the base and pinned at the top. Using Euler’s formula, find its critical buckling load Pcr.
A solid round column of length L made of steel is pinned at both ends. It must carry 130 kN with a factor of safety of 2.5 against buckling. Find the smallest diameter d.
A symmetric I-section has flanges 150 mm wide and 16 mm thick, an overall depth of 280 mm and a web 12 mm thick. Find its elastic section modulus Sx for bending about the horizontal centroidal axis x.
A 150 mm × 200 mm plate has a rectangular cut-out 50 mm wide and 40 mm high, centred on the vertical centre line with its centre 120 mm above the bottom. Find the second moment of area Ix about the horizontal axis through the centroid.
For the soil profile shown, find the vertical effective stress σ′ at point X.
For the soil profile shown, find the vertical effective stress σ′ at point X.
A 7 m thick overconsolidated clay layer has Cc = 0.3, Cr = 0.06 and e₀ = 0.95. At its mid-depth the effective vertical stress is 110 kPa and the preconsolidation pressure is 140 kPa. A new building adds 80 kPa of vertical stress at mid-depth. Find the primary consolidation settlement S in mm.
In a laboratory oedometer test, a clay specimen 25 mm thick, drained at the top and bottom, reaches 50% consolidation in 6 min. In the field the same clay forms a layer 7.5 m thick. It lies between two sand layers, so water drains out at the top and the bottom. How many days t does the field layer take to reach 50% consolidation?
A vertical rectangular gate 3 m wide sits in the wall of a tank of water, with its top edge 2.5 m below the surface. Find the resultant hydrostatic force F on the gate.
A vertical rectangular gate 2 m wide sits in the wall of a tank of oil, with its top edge 2 m below the surface. Find the depth ycp of the centre of pressure below the surface.
Water flows at 25 L/s through a pipe 800 m long with an inside diameter of 100 mm. Use the Darcy friction factor f = 0.028 and g = 9.81 m/s². Find the friction head loss hf.
Two open reservoirs are joined by a pipe 200 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.025. The water surface of the upper reservoir is 28 m above that of the lower one. The minor losses are a square-edged entrance (K = 0.5), 4 bends (K = 0.3 each), a valve (K = 5) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². Find the flow rate Q in L/s.
A concrete storm sewer with an inside diameter of 300 mm flows just full (as an open channel, not under pressure). Manning n = 0.013 and the slope is S = 0.002. Find the flow rate Q in m³/s.
A concrete sewer with an inside diameter of 250 mm (Manning n = 0.013) must reach a self-cleansing velocity of 0.9 m/s when flowing just full. Find the minimum slope S (in m/m).
A pump delivers 100 L/s of water (ρ = 1000 kg/m³) against a head of 38 m. The pump efficiency is 80% and the electric motor driving it has an efficiency of 95%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
A pump moves 75 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 7 m higher. The pipe is 600 m long with an inside diameter of 200 mm and a Darcy friction factor f = 0.018; the minor-loss coefficients add up to ΣK = 5 (entrance and exit included). Use g = 9.81 m/s². Find the head H the pump must add.
An annuity pays €2 000 at the end of each year for 6 years. The interest rate is 10% per year. Find the present value PV of the annuity.
You save €300 at the end of every month for 10 years. The account pays 3% per year, compounded monthly. How much have you saved at the end? Find FV.
A machine costs €40 000 today. It generates €6 500 at the end of each year for 6 years and can then be sold for €6 000. The cost of capital is 8%. Find the net present value NPV.
A project costs €20 000 and returns €13 500, €11 500 at the end of years 1 to 2. Find its internal rate of return IRR in percent.
X is normal with mean 60 s and standard deviation 5 s; Y is independent and normal with mean 45 s and standard deviation 4 s. Find the probability P that X + Y is more than 99.9 s. Round z to two decimals.
Hole diameters D are normal with mean 25.02 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.02 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.
Answer: RA = 34 kN
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Taking moments about one support removes that support’s reaction from the equation.
Replace the supports with reaction forces. A roller gives one vertical reaction; with only vertical loads the pin gives one vertical reaction too.
Moments about A, counter-clockwise positive. Then vertical force balance.
Answer: RA = 29.29 kN
Static equilibrium: a body at rest has ΣFx = 0, ΣFy = 0 and ΣM = 0 about any point. Replace each distributed load by its resultant (its area) acting at the centroid of the load diagram, and take moments about a support to remove that support’s reaction from the equation.
Answer: FBC = −14.14 kN
Method of joints: every pin of a truss is in equilibrium, so ΣFx = 0 and ΣFy = 0 at each joint. Start at a joint with at most two unknown member forces. A negative result means the member is in compression.
Equilibrium of the whole truss. Moments about A, counter-clockwise positive.
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
Answer: FAC = 23.33 kN
Method of joints: every pin of a truss is in equilibrium, so ΣFx = 0 and ΣFy = 0 at each joint. Start at a joint with at most two unknown member forces. A negative result means the member is in compression.
Equilibrium of the whole truss. Moments about A, counter-clockwise positive.
The roller gives only a vertical reaction, so joint C has just two unknown member forces. Assume both in tension.
Answer: Mx = 61.88 kN·m
Method of sections: cut the beam where you want the internal forces and keep one side. The shear force V balances the vertical forces on that side, and the bending moment M balances their moments about the cut.
Equilibrium of the whole beam.
Only forces left of the cut appear. A UDL portion acts as one force at the middle of that portion.
Answer: D (x = 6 m)
Largest bending moment: M changes slope wherever the shear force changes, so its peaks sit where the shear force crosses zero: under a point load, inside a distributed load, or over a support of an overhang. At a fixed support of a cantilever the moment is largest at the wall. Compare the size of M at those places, sagging or hogging.
Take moments about one support, then use the vertical force balance.
Sum the moments of all forces on one side of the cut. Sagging is positive, hogging negative.
Answer: δP = 10.25 mm
Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).
Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.
Answer: hreq = 208.7 mm
Beam deflection: deflections come from integrating the curvature M/(EI) twice. Standard cases are tabulated, for example PL³/(3EI) for a cantilever with an end load and 5wL⁴/(384EI) for a simply supported beam under a full uniform load. Deflections from separate loads add (superposition).
Work in N and mm: P in N, w in N/mm (kN/m = N/mm), L in mm, E in N/mm² (GPa × 1000), I in mm⁴.
In practice, choose the next standard depth above this value.
Answer: Pcr = 19.83 kN
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
A rectangle buckles about its weak axis, using the smaller dimension cubed.
Answer: d = 67.67 mm
Euler buckling: a slender column becomes unstable at Pcr = π²EI/(KL)², where KL is the effective length: K = 1 for pinned–pinned, 2 for fixed–free, 0.7 for fixed–pinned and 0.5 for fixed–fixed. It buckles about the axis with the smallest I. The formula holds for slender columns, whose critical stress stays below the yield stress.
Check: the slenderness KL/r is 147.8, so the column is slender and Euler applies.
Answer: Sx = 707.1 × 10³ mm³
Second moment of area: for a rectangle about its own centroidal axis I = bh³/12, and about its base bh³/3 (parallel-axis theorem I = Ic + A·d²). A hole is handled by subtracting its I, as long as both parts share the same axis. The elastic section modulus is S = I/c, where c is the distance to the extreme fibre.
Both cut-outs together are (B − tw) wide and hw = H − 2tf high, and share the axis x.
Answer: Ix = 98.88 × 10⁶ mm⁴
Sections with holes: treat a hole as a negative area. The centroid is ȳ = (A₁y₁ − A₂y₂)/(A₁ − A₂), with A₁, y₁ for the full plate and A₂, y₂ for the hole. For the second moment of area about the centroidal axis, use the parallel-axis theorem for each part and subtract the hole: I = (Ic1 + A₁d₁²) − (Ic2 + A₂d₂²), where d is the distance from each part's own centroid to the axis through ȳ. A rectangle has Ic = bh³/12 and a circle Ic = πd⁴/64.
Answer: σ′ = 93.57 kPa
Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).
Add the weight of each soil band above X. Above the water table use the dry unit weight, below it the saturated one.
Water pressure grows with depth below the water table.
Answer: σ′ = 75.66 kPa
Effective stress (Terzaghi): the soil skeleton carries σ′ = σ − u. The total stress σ is the weight of everything above the point, Σγ·h, using the dry unit weight above the water table and the saturated unit weight below it. Below the water table the pore water pressure is hydrostatic: u = γw·(depth below the water table).
Add the weight of each soil band above X. Above the water table use the dry unit weight, below it the saturated one.
Water pressure grows with depth below the water table.
Answer: S = 165.4 mm
Primary consolidation of a clay layer of thickness H, using the effective stress at the middle of the layer: σ′f = σ′0 + Δσ. Normally consolidated clay: S = Cc·H/(1 + e0)·log₁₀(σ′f/σ′0). Overconsolidated clay recompresses along Cr up to the preconsolidation pressure σ′p and follows Cc only beyond it. Time: t = Tv·Hdr²/cv, where the drainage path Hdr is H/2 when water can leave at the top and bottom, and H when only one side drains.
Recompression up to σ′p, virgin compression beyond it.
Answer: t = 375 days
Primary consolidation of a clay layer of thickness H, using the effective stress at the middle of the layer: σ′f = σ′0 + Δσ. Normally consolidated clay: S = Cc·H/(1 + e0)·log₁₀(σ′f/σ′0). Overconsolidated clay recompresses along Cr up to the preconsolidation pressure σ′p and follows Cc only beyond it. Time: t = Tv·Hdr²/cv, where the drainage path Hdr is H/2 when water can leave at the top and bottom, and H when only one side drains.
t = Tv·Hdr²/cv with the same T_v and cv in the lab and the field.
Answer: F = 275.9 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Answer: ycp = 3.41 m
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Pressure grows with depth, so the resultant acts below the centroid.
Answer: hf = 115.7 m
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Answer: Q = 128.8 L/s
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Both surfaces are open to the air and practically at rest, so all of Δz is lost in the pipe.
Answer: Q = 0.04325 m³/s
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
Answer: S = 0.005519 m/m
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
That is a fall of 5.519 m per kilometre.
Answer: Pel = 49.05 kW
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Answer: H = 24.14 m
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Both surfaces are open and practically at rest, so pressure and velocity terms drop out.
Answer: PV = 8711 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
Answer: FV = 41922 €
Time value of money: money today is worth more than the same amount later, because it can earn interest. With interest rate i per period, an amount grows as FV = PV·(1 + i)n. A series of equal payments (an annuity) is valued by adding up each payment, which gives the annuity formulas.
You paid in €36 000 yourself; the rest is interest.
Answer: NPV = −6170 €
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
Answer: IRR = 16.75 %
Net present value: discount every cash flow back to today with the required rate of return r, then add them up: NPV = Σ CFₜ/(1 + r)ᵗ. A project with NPV > 0 earns more than r and adds value. The internal rate of return (IRR) is the rate at which NPV = 0.
NPV falls as the rate rises, so the IRR lies between a rate with positive NPV and one with negative NPV.
Linear interpolation gives a first estimate; repeating it (or a calculator's IRR function) gives the exact rate.
Answer: P = 0.7881
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
Answer: P = 0.8133
Sums and differences of normal variables: if X and Y are independent and normal, X + Y and X − Y are normal too. Means add or subtract: E(X ± Y) = μX ± μY. Variances always add, also for a difference: Var(X ± Y) = σX² + σY², so σ = √(σX² + σY²). Standard deviations never add directly. The total of n independent items with the same distribution has mean n·μ and standard deviation σ·√n. Then standardise as usual: z = (x − mean)/σ and read Φ(z).
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